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According to a certain estimate, the depth N(t), in centimeters, of
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23 Jan 2012, 22:37
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According to a certain estimate, the depth N(t), in centimeters, of the water in a certain tank at t hours past 2:00 in the morning is given by N(t)= 20(t  5)^2 + 500 for 0 ≤ t ≤ 10. According to this estimate, at what time in the morning does the depth of the water in the tank reach its maximum? a) 5:30 b) 7:00 c) 7:30 d) 8:00 e) 9:00
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Re: According to a certain estimate, the depth N(t), in centimeters, of
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24 Jan 2012, 02:45
Baten80 wrote: According to a certain estimate, the depth N(t), in centimeters, of the water in a certain tank at t hours past 2:00 in the morning is given by N(t)= 20(t5)^²+500 for 0≤t≤10. According to this estimate, at what time in the morning does the depth of the water in the tank reach its maximum?
a) 5:30 b) 7:00 c) 7:30 d) 8:00 e) 9:00 Don't get bogged down by the dirty N(t) expression. Just think of it this way: N(t) is a combination of two terms: a positive term (500) and a negative term (\(20(t5)^2\)). To maximize N(t), I need to make the positive term as large as possible (It is a constant here so I cannot do much with it) and the absolute value of the negative term as small as possible. The smallest absolute value is 0. Can I make it 0? Yes, if I make t = 5, the negative term becomes 0 and N(t) is maximized. My answer must be 2:00 + 5 hrs i.e. 7:00. Most of the maximum minimum questions on GMAT will require you to only think logically. The calculations involved will be minimum.
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Re: According to a certain estimate, the depth N(t), in centimeters, of
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23 Jan 2012, 22:44
the expression is N(t)= 20(t5)^²+500 (of course valid after 2:00 in the morning)
"the depth would be maximum" means the value of the above expression should be maximum or the value of square term (which has a negative 20 attached to it) should be minimum i.e. zero
the square part is zero at t=5
so the time at which the depth is maximum is 2:00 + 5 hrs= 7:00 (B)



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Re: According to a certain estimate, the depth N(t), in centimeters, of
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23 Jan 2012, 22:52
puneetkr wrote: the expression is N(t)= 20(t5)^²+500 (of course valid after 2:00 in the morning)
"the depth would be maximum" means the value of the above expression should be maximum or the value of square term (which has a negative 20 attached to it) should be minimum i.e. zero
the square part is zero at t=5
so the time at which the depth is maximum is 2:00 + 5 hrs= 7:00 (B) if 20(t5)^²=0 then t = 5 But Why is the 500 of the equation is not considered? Show your complete calculation.
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Re: According to a certain estimate, the depth N(t), in centimeters, of
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23 Jan 2012, 23:01
this is the complete calculation
N(t)= 20(t5)^²+500 now for any value of t we put in; we get some negative value of 20(t5)^² (say x) so our expression is now N(t)=500x
this expression would have maximum value only when x is minimum we know the minimum value for a square term is "zero" and (x has a square term) and that comes when t=5 i.e. when we put t=5 here we get N(t) = 5000 = 500



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Re: According to a certain estimate, the depth N(t), in centimeters, of
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24 Jan 2012, 01:35
Baten80 wrote: puneetkr wrote: the expression is N(t)= 20(t5)^²+500 (of course valid after 2:00 in the morning)
"the depth would be maximum" means the value of the above expression should be maximum or the value of square term (which has a negative 20 attached to it) should be minimum i.e. zero
the square part is zero at t=5
so the time at which the depth is maximum is 2:00 + 5 hrs= 7:00 (B) if 20(t5)^²=0 then t = 5 But Why is the 500 of the equation is not considered? Show your complete calculation. According to a certain estimate, the depth N(t), in centimeters, of the water in a certain tank at t hours past 2:00 in the morning is given by N(t)= 20(t5)^²+500 for 0≤t≤10. According to this estimate, at what time in the morning does the depth of the water in the tank reach its maximum?A. 5:30 B. 7:00 C. 7:30 D. 8:00 E. 9:00 Consider this: \(20(t5)^2\leq{0}\) hence \(50020(t5)^2\) reaches its maximum when \(20(t5)^2=0\), thus when \(t=5\). 2:00AM+5 hours=7:00AM. Answer: B. Hope it helps.
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Re: According to a certain estimate, the depth N(t), in centimeters, of
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Re: According to a certain estimate, the depth N(t), in centimeters, of
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27 Nov 2015, 22:58
Can someone please explain why A is not the answer?
Because if I plug in t = 3.5 (which is 5.30 (3.5 after the 2:00) , it gives me the maximum value of 20(t5)^2 expression, which will be positive. Hence, the height will be at its max of 545. Hence, shouldn't the answer be A?
Please explain the reasoning.
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Re: According to a certain estimate, the depth N(t), in centimeters, of
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28 Nov 2015, 08:33
herein wrote: Can someone please explain why A is not the answer?
Because if I plug in t = 3.5 (which is 5.30 (3.5 after the 2:00) , it gives me the maximum value of 20(t5)^2 expression, which will be positive. Hence, the height will be at its max of 545. Hence, shouldn't the answer be A?
Please explain the reasoning.
Thank you Please read the solutions above. Also, if t = 3.5, then 500  20(3.5  5)^2 = 455, not 545.
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Re: According to a certain estimate, the depth N(t), in centimeters, of
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28 Nov 2015, 09:25
Baten80 wrote: According to a certain estimate, the depth N(t), in centimeters, of the water in a certain tank at t hours past 2:00 in the morning is given by N(t)= 20(t  5)^2 + 500 for 0 ≤ t ≤ 10. According to this estimate, at what time in the morning does the depth of the water in the tank reach its maximum?0 N(t)= \(20(t  5)^2\) + 500 ; The red highlighted portion of the equation is the most critical part, further t = 2am + time past 2amNow why \(20(t  5)^2\) is important ; Any value of t less than 5 will result in a negative value of \(20(t  5)^2\) Lets check : Time : 3am , t = 1 { t = 2am + time past 2am }\(20(t  5)^2\) \(20(1  5)^2\) \(20(4)^2\) \(20(16)\) \(320\) When put in the final equation : \(20(t  5)^2\) the result will be 320 + 500 => 180 Check a few more any value of t less than 5 will result in ve value of \(20(t  5)^2\) and will ultimately lead to a depth of water less than 500.Our target is getting t = 5 ; since at t = 5
\(20(t  5)^2\) will be 0 and the final equation \(20(t  5)^2\) + 500 will be maximum ; ie 500We already know t = 2am + time past 2am So, 7 = 2am + 5 hours. Answer is definitely (B)
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Re: According to a certain estimate, the depth N(t), in centimeters, of
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28 Nov 2015, 16:07
Hi All, These specific types of "limit" questions are relatively rare on Test Day, although you'll likely be tested on the concept at least once. Whenever you're asked to minimize or maximize a value, you should look to do something with the other "pieces" of the equation (usually involving maximizing or minimizing those pieces). In the given equation, notice how you have two "parts": the 20(something) and a +500. Here, to MAXIMIZE the value of N(t), we have to minimize the "impact" that the 20(something) has on the +500. By making that first part equal 0, we'll be left with 0 + 500. Mathematically, we have to make whatever is inside the parentheses equal 0.... (T5) = 0 T = 5 Since T represents the number of hours past 2:00am, we know that at 7:00am, the water will reach 500cm (the maximum value). Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: According to a certain estimate, the depth N(t), in centimeters, of
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14 Apr 2018, 17:49
A useful trick to solve Maximum of any equation,with a single vraiable:
Just differentiate the equation with the variable and equate it to zero to find the solution. That solution is the value at which the euqation will have optimum value.
N(t)= 20(t5)^²+500
d/dt N(t)= 40(t5)=0
40t=200 => t=5
So, 7 = 2am + 5 hours.



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Re: According to a certain estimate, the depth N(t), in centimeters, of
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04 Jun 2018, 03:38
valardohaeris wrote: A useful trick to solve Maximum of any equation,with a single vraiable:
Just differentiate the equation with the variable and equate it to zero to find the solution. That solution is the value at which the euqation will have optimum value.
N(t)= 20(t5)^²+500
d/dt N(t)= 40(t5)=0
40t=200 => t=5
So, 7 = 2am + 5 hours. Hi, I used that method too. I would like to know if there are always questions like this on every exam. BR Ge



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Re: According to a certain estimate, the depth N(t), in centimeters, of
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05 Jun 2018, 10:32
Baten80 wrote: According to a certain estimate, the depth N(t), in centimeters, of the water in a certain tank at t hours past 2:00 in the morning is given by N(t)= 20(t  5)^2 + 500 for 0 ≤ t ≤ 10. According to this estimate, at what time in the morning does the depth of the water in the tank reach its maximum?
a) 5:30 b) 7:00 c) 7:30 d) 8:00 e) 9:00 Since 20(t  5)^2, will be a nonpositive number, its maximum value is 0 when t = 5, and the maximum value of the function will then be: N(5) = 20(5  5)^2 + 500 = 500 Thus, the maximum depth is at 2am + 5 hours = 7am. Answer: B
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According to a certain estimate, the depth N(t), in centimeters, of
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14 Feb 2019, 07:05
Can someone explain why / how you would know not to simplify the original provided equation (20(t5)^2 + 500)? When you simplify the equation, you seems to get a different answer. Please see work below:
20(t5)^2 + 500 20(t^2  10t + 25) + 500 < foil 20t^2 + 200t  500 + 500 < distributed the 20 20t^2 + 200t 20t(t + 10t) < simplified formula
Given the 20t(t+10t) is simplified correctly, then the water level will continue to grow with every hour with the latest hour being the maximum. This is different than the answer provided, where t is at the max once 5 hours past.



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Re: According to a certain estimate, the depth N(t), in centimeters, of
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14 Feb 2019, 14:42
rjacobson99 wrote: Can someone explain why / how you would know not to simplify the original provided equation (20(t5)^2 + 500)? When you simplify the equation, you seems to get a different answer. Please see work below:
20(t5)^2 + 500 20(t^2  10t + 25) + 500 < foil 20t^2 + 200t  500 + 500 < distributed the 20 20t^2 + 200t 20t(t + 10t) < simplified formula
Given the 20t(t+10t) is simplified correctly, then the water level will continue to grow with every hour with the latest hour being the maximum. This is different than the answer provided, where t is at the max once 5 hours past. Hi rjacobson99, In the last "step" of you work, you have not properly accounted for the 'minus' sign. If you want to factor out "20t", then that's fine, but here's what you would be left with: 20t^2 + 200t (20t)(t + 10) (20t)(10  t) The maximum result will occur when t = 5. GMAT assassins aren't born, they're made, Rich
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According to a certain estimate, the depth N(t), in centimeters, of
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15 Feb 2019, 09:07
EMPOWERgmatRichC wrote: rjacobson99 wrote: Can someone explain why / how you would know not to simplify the original provided equation (20(t5)^2 + 500)? When you simplify the equation, you seems to get a different answer. Please see work below:
20(t5)^2 + 500 20(t^2  10t + 25) + 500 < foil 20t^2 + 200t  500 + 500 < distributed the 20 20t^2 + 200t 20t(t + 10t) < simplified formula
Given the 20t(t+10t) is simplified correctly, then the water level will continue to grow with every hour with the latest hour being the maximum. This is different than the answer provided, where t is at the max once 5 hours past. Hi rjacobson99, In the last "step" of you work, you have not properly accounted for the 'minus' sign. If you want to factor out "20t", then that's fine, but here's what you would be left with: 20t^2 + 200t (20t)(t + 10) (20t)(10  t) The maximum result will occur when t = 5. GMAT assassins aren't born, they're made, Rich Hi Rich, In the correct version of the formula (20t)(10t), is there a way to determine the inflection point of t = 5 without having to plug in values from t = 1 to 6? I.e. is there a shortcut? RJ



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Re: According to a certain estimate, the depth N(t), in centimeters, of
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15 Feb 2019, 12:55
rjacobson99 wrote: EMPOWERgmatRichC wrote: rjacobson99 wrote: Can someone explain why / how you would know not to simplify the original provided equation (20(t5)^2 + 500)? When you simplify the equation, you seems to get a different answer. Please see work below:
20(t5)^2 + 500 20(t^2  10t + 25) + 500 < foil 20t^2 + 200t  500 + 500 < distributed the 20 20t^2 + 200t 20t(t + 10t) < simplified formula
Given the 20t(t+10t) is simplified correctly, then the water level will continue to grow with every hour with the latest hour being the maximum. This is different than the answer provided, where t is at the max once 5 hours past. Hi rjacobson99, In the last "step" of you work, you have not properly accounted for the 'minus' sign. If you want to factor out "20t", then that's fine, but here's what you would be left with: 20t^2 + 200t (20t)(t + 10) (20t)(10  t) The maximum result will occur when t = 5. GMAT assassins aren't born, they're made, Rich Hi Rich, In the correct version of the formula (20t)(10t), is there a way to determine the inflection point of t = 5 without having to plug in values from t = 1 to 6? I.e. is there a shortcut? RJ Hi rjacobson99, Unfortunately, manipulating the equation in the way that you did places the variable "t" in both pairs of parentheses, so there isn't an 'obvious' solution to maximize the value. However, in the original equation, there IS an obvious Number Property that you can use... N(t) = 20(t  5)^2 + 500 for 0 ≤ t ≤ 10. In the given equation, notice how you have two "parts": the 20(something) and a +500. Here, to MAXIMIZE the value of N(t), we have to minimize the "impact" that the negative term  the 20(something)  has on the +500. By making that first part equal 0, we'll be left with 0 + 500. Mathematically, we have to make whatever is inside the parentheses equal 0.... (T5) = 0 T = 5 GMAT assassins aren't born, they're made, Rich
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